Acceleration in 2 Dimensions-Vectors and Projectiles

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An airplane is initially flying at a velocity of (150, 0) m/s and is subjected to accelerations of 8.92 m/s² in the x-direction and -1.21 m/s² in the y-direction. To find the airplane's velocity after 3.6 seconds, the change in velocity for each component must be calculated using the formula Δv = vf - vi. The calculations yield a change of 32.1 m/s in the x-direction and 4.4 m/s in the y-direction. The final velocity is determined by adding these changes to the initial velocity components, resulting in two distinct answers for the x and y components.
Snape1830
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1. An airplane is flying with a velocity of (150, 0) m/s. It is accelerated at 8.92 m/s2 in the x direction, and accelerated in negative y direction at 1.21 m/s2. What is the airplane's velocity at t = 3.6 s?2. My textbook says this is a relevant equation: average acceleration = change in velocity/change in time.
ax=change in velocityx/change in time
ay=change in velocityy/change in time.

Where vx and vy are the components of velocity.
3. So, I plugged the knows (time and acceleration) into the equations and got 32.1 m/s and 4.4 m/s. But I knew it was wrong because why would they give (150, 0) m/s?

How do I solve this problem?
 
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You are finding the change in the velocity components. Use Δv = vf - vi.
 
Thanks Doc Al. The only issue is that I need two answers. (x,y) m/s.
 
Snape1830 said:
The only issue is that I need two answers. (x,y) m/s.
And you have two components. What's the initial velocity x-component? y-component?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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